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A276323
a(n) = (binomial(2 * prime(n + 3), prime(n + 3)) * A005259(prime(n + 3) - 1) - 2)/prime(n + 3)^5 for n >= 1.
2
4382314, 59821998476834, 338197165389273486, 17314015796594772560245514, 145853326344012138627669357202, 12936469013977571458378002436843685186, 15931675838688077485749893663903436780403973163302
OFFSET
1,1
COMMENTS
Let p be a prime > 5. Binomial(2 * p, p) * A005259(p - 1) == 2 (mod p^5). So a(n) is an integer.
LINKS
Julian Rosen, Periods, supercongruences, and their motivic lifts, arXiv:1608.06864 [math.NT], 2016.
EXAMPLE
a(1) = (binomial(14, 7) * A005259(6) - 2)/7^5 = (3432 * 21460825 - 2)/7^5 = 4382314.
MATHEMATICA
Table[(Binomial[2 Prime[n + 3], Prime[n + 3]] Sum[(Binomial[#, k] Binomial[# + k, k])^2, {k, 0, #}] &[Prime[n + 3] - 1] - 2)/Prime[n + 3]^5, {n, 7}] (* Michael De Vlieger, Aug 30 2016 *)
PROG
(Ruby)
require 'prime'
def C(n, r)
r = [r, n - r].min
return 1 if r == 0
return n if r == 1
numerator = (n - r + 1..n).to_a
denominator = (1..r).to_a
(2..r).each{|p|
pivot = denominator[p - 1]
if pivot > 1
offset = (n - r) % p
(p - 1).step(r - 1, p){|k|
numerator[k - offset] /= pivot
denominator[k] /= pivot
}
end
}
result = 1
(0..r - 1).each{|k|
result *= numerator[k] if numerator[k] > 1
}
return result
end
def A005259(n)
i = 0
a, b = 1, 5
ary = [1]
while i < n
i += 1
a, b = b, ((((34 * i + 51) * i + 27) * i + 5) * b - i ** 3 * a) / (i + 1) ** 3
ary << a
end
ary
end
def A276323(n)
p_ary = Prime.take(n + 3)[3..-1]
a = A005259(p_ary[-1] - 1)
ary = []
p_ary.each{|i|
j = C(2 * i, i) * a[i - 1] - 2
break if j % i ** 5 > 0
ary << j / i ** 5
}
ary
end
CROSSREFS
Sequence in context: A254259 A254305 A237537 * A288086 A210297 A019288
KEYWORD
nonn
AUTHOR
Seiichi Manyama, Aug 30 2016
STATUS
approved